ASEN 5022 - Spring 2005Dynamics of Aerospace StructuresLecture 06: January 27Energy Principles for DynamicsPrinciple of Virtual Work for StaticsFor statics a la Meirovitch:Ni=1Fi· δri= 0 (1)where{Fi} are impressed forces, and{δ} designates the virtual character of the instan-taneous variations, as opposed to the differentialsymbol {d} designating actual differentials of po-sitions {ri} taking place in the time interval {dt}.Principle of Virtual Work for DynamicsFor dynamics a la d’Alembert:Ni=1(Fi− mi¨ri) · δri= 0 (2)where{¨ri} is the acceleration of particle mi.Generalized CoordinatesVirtual displacements δrimay be expressed interms of the generalized virtual displacementsδqk(k = 1, 2, 3, ..., n):δri=nk=1∂ri∂qkδqk, i = 1, 2, 3, ..., N (3)An example:r = xi + yj + L(cos θi + sinθj)(4)Generalized Coordinates-cont’dObserve that there are three independent variables,(x, y,θ).δr =∂r∂xδx +∂r∂yδy +∂r∂θδθ= δxi + δyj + L(−sinθi +cosθj)δθ(5)Note that the dimensional unit of (x, y) is meter(m)whereas that of θ is radian, suggesting that the unitsof generalized coordinates can be different.Interpretation of Equation (2)Let’s revisit the formulation of the equations of mo-tion for the spring-mass-bar problem studied in theprevious lecture. To this end, first, wemust obtain thevirtual displacements.A Spring-Mass-Bar System:xmKoKomθx,iLMgFFACy, jBB‘C‘A‘Nom x’’NAθkxmgA‘ACXACYθMgFM r’’Cθ”ML2/12)(B‘C‘A‘ACXACYVirtual displacements:Virtual displacement for mass m:δrA= δx i (6)Virtual displacement for bar M:δrC= δrA+ δrACδrAC=L2(cosθi + sinθj)δθ(7)Virtual displacements-cont’dVirtual rotation for bar M:Here, one utilizes the angular velocityω =˙θktoobtain the virtual rotationδθ = δθk (8)In general one hasδθ = δθxi + δθyj + δθzk (9)Acceleration Vectors:Acceleration vector for mass m:aA=˙vA=¨xi (10)Acceleration vector for bar M:aC= aA+ aACaAC=L¨θ2(cosθi + sinθj)−L˙θ22(sinθi − cosθj)(11)Equilibrium equation for mass m:om x’’NAθk xmgA‘ACXACYfA=−kxi − m ¨xi − mgj + NAj+XACi + YACj = 0 (12)Equilibrium equation for bar M:θMgFM r’’Cθ”ML2/12)(B‘C‘A‘ACXACYfC= Fi − MaC− Mgj − XACi − YACj = 0 (13)MC=−ML2¨θ12k + rCA× (−XACi − YACj)+rCB× Fi = 0 (14)An Application of D’Alembert’s Principle for thespring-mass-bar problem(Fi− mi¨ri) · δri=fA· δrA+ fC· δrC+ MC· δθC= 0⇓(15)Note that fA· δrA= 0 and so other two terms!{−kxi − m ¨xi − mgj + NAj+ XACi + YACj}·δrA+{Fi − MaC− Mgj− XACi − YACj}·δrC{−ML2¨θ12k + rCB× Fi+ rCA× (−XACi − YACj)}·δθ(16)The first term in (16) becomesfA· δrA={−kxi − m ¨xi − mgj + NAj+ XACi + YACj}·(δxi)fA· δrA={−kx − m ¨x + XAC} δx(17)Remark: Observe that the reactions forces YACand NAhave played no role in the resulting virtual work!The second term in (16) becomesfC· δrC={Fi − MaC− Mgj − XACi − YACj}·δrC={Fi − M[¨xi +L¨θ2(cosθi + sin θj)−L˙θ22(sinθi − cos θj)]− Mgj − XACi − YACj}··{δxi +L2(cosθi + sin θj)δθ}(18)fC· δrC={F − M[¨x +L¨θ2cosθ −L˙θ22sinθ]− XAC} δx+{F − M[¨x +L¨θ2cosθ −L˙θ22sinθ]− XAC}L2cosθδθ+{−M[L¨θ2sinθ −L˙θ22cosθ] − Mg− YAC}L2sinθδθ(19)fC· δrC={F − M[¨x +L¨θ2cosθ −L˙θ22sinθ]− XAC} δx+{L cosθ2F −ML2cosθ ¨x −ML24¨θ−L2cosθ XAC−MLg2sinθ−L2sinθYAC} δθ(20)For the third term evaluation, compute:rCB× Fi =L2(sinθi − cos θj) × Fi=FL2cosθk(21)rCA× (−XACi − YACj) ={L2cosθ XAC+L2sinθYAC}k(22)Finally, we obtain:MC· δθC={−ML2¨θ12+FL2cosθ+ [L2cosθ XAC+L2sinθYAC]}δθ(23)Combine (17), (20) and (23) to obtain{(M + m) ¨x + kx +ML2cosθ¨θ−ML2sinθ˙θ2− F} δx+{ML23¨θ ¨x +ML2cosθ ¨x+MgL2sinθ − L cosθ F}δθ = 0(24)Since δx and δθ are arbitrary, we obtain:(M + m) ¨x + kx +ML2cosθ¨θ−ML2sinθ˙θ2= F(25)ML23¨θ +ML2cosθ ¨x +MgL2sinθ= L cosθ F(26)Observations1. When using Newton’s second law, Eqs.(25) and (26)are obtained by eliminating the reaction forces XACand YAC.2. On the other hand, one does not need to considerreaction forces in applying d’Alembert’s principle!Only apparent forces including inertia forces need tobe considered. This is a major advantage of applyingenergy principles over Newton’s